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Product And Quotient Rules

The product rule states (fg)′=f′g+fg′(fg)' = f'g + fg' and the quotient rule states (fg)′=f′g−fg′g2(\frac{f}{g})' = \frac{f'g - fg'}{g^2}. Both are essential because derivatives of products and quotients cannot be found by simply differentiating each piece separately.

Product and Quotient Rules

When differentiating a product or quotient of functions, you cannot simply differentiate each piece separately. Instead, use these two fundamental rules.

The Rules

If h(x)=f(x)g(x)h(x) = f(x)g(x), then the product rule is: h′(x)=f′(x)g(x)+f(x)g′(x)h'(x) = f'(x)g(x) + f(x)g'(x)

If h(x)=f(x)g(x)h(x) = \frac{f(x)}{g(x)}, then the quotient rule is: h′(x)=f′(x)g(x)−f(x)g′(x)[g(x)]2h'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}

Why These Rules Matter

The intuitive guess that (fg)′=f′g′(fg)' = f'g' fails. For example, if f(x)=xf(x) = x and g(x)=xg(x) = x, then fg=x2fg = x^2 so (fg)′=2x(fg)' = 2x, but f′g′=1⋅1=1f'g' = 1 \cdot 1 = 1. The product rule exists because both functions change simultaneously—the rate of change of a product depends on how each piece contributes to the overall change.

Worked Examples

Product Rule: Differentiate h(x)=x2sin⁡xh(x) = x^2 \sin x.
Let f(x)=x2f(x) = x^2 and g(x)=sin⁡xg(x) = \sin x, so f′(x)=2xf'(x) = 2x and g′(x)=cos⁡xg'(x) = \cos x. h′(x)=2xsin⁡x+x2cos⁡xh'(x) = 2x\sin x + x^2 \cos x

Quotient Rule: Differentiate h(x)=x2sin⁡xh(x) = \frac{x^2}{\sin x}.
Using the same ff and gg: h′(x)=2xsin⁡x−x2cos⁡xsin⁡2xh'(x) = \frac{2x\sin x - x^2\cos x}{\sin^2 x}

Note the order and minus sign—these are critical differences from the product rule.

Common Pitfall

A helpful mnemonic for the quotient rule is "low d-high minus high d-low, over low squared," where "low" is the denominator and "high" is the numerator. This prevents accidentally reversing the subtraction, which would flip your answer's sign.

Combining Rules

When pieces are composite functions, combine the product or quotient rule with the chain rule. For x2e3xx^2 e^{3x}: apply the product rule where f′=2xf' = 2x and g′=3e3xg' = 3e^{3x} (using the chain rule on the exponential), yielding h′(x)=2xe3x+3x2e3xh'(x) = 2xe^{3x} + 3x^2e^{3x}.

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